删除最多 P 条边后检查每个连接组件的 XOR 是否相等

给定一棵具有N个节点和一个整数P的树，任务是删除范围 [1, P) 内的边，并为形成的每个连接组件找到节点的 XOR。如果节点的值对于所有形成的连接组件都相等，则打印“YES”否则“NO”。

例子：

Input: N = 5, P = 5, Edges[][]={ { 1, 2 }, { 2, 3 }, { 1, 4 }, { 4, 5 }}, nodes[] = {2, 2, 2, 2, 2}

Output: YES
Explanation: We can remove all edges, then there will be 5 connected components each having one node with value 2, thus XOR of each of them will be 2.
 

Input: N = 5,  P = 2, Edges[][]={ { 1, 2 }, { 2, 3 }, { 1, 4 }, { 4, 5 }}, nodes[] = {1, 6, 4, 1, 2}

Output: YES
Explanation: As, p = 2, atleast one edge need to be removed so, removing edge (4, 5), gives two connected components.
XOR of first component would be, arr₁ XOR  arr₂ XOR arr₃ XOR arr₄= 1 XOR 6 XOR 4 XOR 1 => 2.
XOR of second component will be arr₅ = 2. (Thus, both equal).

方法：解决此问题的方法基于以下观察：

The fact here, is that deletion of edges should be either once or twice, as:

• Removing one edge gives us 2 connected components, which should have same XOR and thus by properties of xor if XOR1 == XOR2, then XOR1 ^ XOR2 = 0, where ^ is the bitwise XOR.
• Similarly, if XOR of the whole tree is 0, removing any edge, will give 2 connected components where both components would have equal value, thus answer = “YES”.
• Now, if we delete 2 edges, deleting more than that would just merge the other components with each other. For ex: if there are 4 or more connected components, we can reduce and merge them, as for 4 it can be reduced to 2 components (which will fall in the case if we remove 1 edge).
• Now, deleting 2 edges gives us a minimum of 3 connected components, using properties of xor we know, that XOR1 ^ XOR2 ^ XOR3 = let’s say some value(y), which means we need to find such subtrees(or say search for 2 edges to remove) whose XOR is equal to some value(y), and if we find it then answer = “YES” else “NO”, such that p>2.
• For ex: say, if we have total XOR = some value(y), with N = 4, there can be 3 segments whose XOR would have been y, then the total elements could get XOR = y, such that p > 2.
  So, overall it can be said that if we find two subtrees whose XOR = y, and if our total XOR was y, we can say that our left out component/segment XOR would also be y, thus answer = “YES”, such that p > 2, using above property XOR1 ^ XOR2 ^ XOR3 = some value(y),

请按照以下步骤应用上述方法：

要找到这样的子树，至少有 3 个连接的组件，可以使用 DFS 遍历轻松完成，并且在回溯期间，我们将在每次迭代时存储每个索引 XOR。

下面是上述方法的实现：

C++

// C++ code for Check after removing
// edges bitwise XOR of each connected
// component formed is equal
 
#include 
using namespace std;
 
const int N = 1e5;
 
// adjacency list to form a tree
vector adj[N];
 
// array arr defined to store node values
// array cal_XOR defined to store
// xor values rooted at i'th level
int nodes[N], cal_XOR[N];
 
// defined a visited array
vector visited(N);
 
// defined a counter to store
// number of subtrees having value equal
// to whole tree XOR
int counter = 0;
 
// first dfs function to find
// xor of subtrees
int dfs_First(int x)
{
    // initializing cal_XOR
    // array with tree node's value
    cal_XOR[x] = nodes[x];
 
    // marking each visited node as true
    visited[x] = true;
 
    // iterating through current node children
    for (auto y : adj[x]) {
        // check if node->child is not visited
        if (!visited[y]) {
            // computing xor of subtree rooted at x
            cal_XOR[x] ^= dfs_First(y);
        }
    }
 
    // returning xor of subtree rooted at x
    return cal_XOR[x];
}
 
// second dfs function to count
// subtrees having values equal
// to entire XOR of tree nodes values
// and returning subtree XOR till that point
int dfs_Second(int x)
{
 
    // marking each visited node as true
    visited[x] = true;
 
    // storing value of nodes[x]
    // to another variable temp
    // for further calculation
    int temp = nodes[x];
 
    // iterating through current node children
    for (auto y : adj[x]) {
 
        // check if node->child is not visited
        if (!visited[y]) {
 
            // storing xor of subtree
            temp ^= dfs_Second(y);
        }
    }
 
    // now, checking if xor of subtree
    // computed till now is equal to
    // entire XOR of tree (root)
    // i.e. value at cal_XOR[0] ->
    // which will give entire XOR of the tree
    if (temp == cal_XOR[0]) {
        // then, make that xor 0
        temp = 0;
 
        // increase count by 1, which
        // means we found a subtree
        counter++;
    }
 
    // return xor of subtree
    // till that point
    return temp;
}
// Function to add edges
void addEdge(int u, int v)
{
    adj[u].push_back(v);
    adj[v].push_back(u);
}
// Function to take input
// for (n-1) edges
void init(int edges[4][2], int numEdges)
{
    for (int i = 0; i < numEdges; i++) {
        edges[i][0]--;
        edges[i][1]--;
        addEdge(edges[i][0], edges[i][1]);
    }
}
 
// Driver Code
int main()
{
 
    // taking input
    int n = 5, p = 2;
    nodes[0] = 1, nodes[1] = 6, nodes[2] = 4,
    nodes[3] = 1, nodes[4] = 2;
 
    // making our visited array false
    for (int i = 0; i < n; i++) {
        visited[i] = false;
    }
 
    // taking input for (n-1) edges
    int edges[4][2]
        = { { 1, 2 }, { 2, 3 }, { 1, 4 }, { 4, 5 } };
    init(edges, 4);
 
    // First dfs Function call
    dfs_First(0);
 
    // again, marking visited array to false
    for (int i = 0; i < n; i++) {
        visited[i] = false;
    }
 
    // initializing answer variable
    bool answer = false;
 
    // if we found XOR of entire tree
    // equal to zero, answer = "YES", as
    // it means there are two connected
    // components equal to each other
    // thus, a single edge can be removed
    if (cal_XOR[0] == 0) {
        answer = true;
    }
    else {
        // second DFS function call
        dfs_Second(0);
 
        // if we found 2 subtree having
        // equal value with XOR of entire tree
        // answer is always "YES", such that
        // p > 2
        if (counter >= 2 and p != 2) {
            answer = true;
        }
    }
    // printing the final answer
    if (answer == true) {
        cout << "YES"
             << "\n";
    }
    else {
        cout << "NO"
             << "\n";
    }
 
    // making counter = 0 for next iteration
    counter = 0;
 
    for (int i = 0; i < n; i++) {
 
        // similarly clearing adjacency list
        // and both arr and cal_XOR array
        adj[i].clear();
        cal_XOR[i] = nodes[i] = 0;
    }
}

Javascript

       // JavaScript code for the above approach
       let N = 1e5;
 
       // adjacency list to form a tree
       let adj = new Array(N);
       for (let i = 0; i < N; i++) {
           adj[i] = [];
       }
 
       // array arr defined to store node values
       // array cal_XOR defined to store
       // xor values rooted at i'th level
       let nodes = new Array(N), cal_XOR = new Array(N);
 
       // defined a visited array
       let visited = new Array(N).fill(false);
 
       // defined a counter to store
       // number of subtrees having value equal
       // to whole tree XOR
       let counter = 0;
 
       // first dfs function to find
       // xor of subtrees
       function dfs_First(x)
       {
        
           // initializing cal_XOR
           // array with tree node's value
           cal_XOR[x] = nodes[x];
 
           // marking each visited node as true
           visited[x] = true;
 
           // iterating through current node children
           for (let y of adj[x])
           {
            
               // check if node->child is not visited
               if (!visited[y])
               {
                
                   // computing xor of subtree rooted at x
                   cal_XOR[x] ^= dfs_First(y);
               }
           }
 
           // returning xor of subtree rooted at x
           return cal_XOR[x];
       }
 
       // second dfs function to count
       // subtrees having values equal
       // to entire XOR of tree nodes values
       // and returning subtree XOR till that point
       function dfs_Second(x) {
 
           // marking each visited node as true
           visited[x] = true;
 
           // storing value of nodes[x]
           // to another variable temp
           // for further calculation
           let temp = nodes[x];
 
           // iterating through current node children
           for (let y of adj[x]) {
 
               // check if node->child is not visited
               if (!visited[y]) {
 
                   // storing xor of subtree
                   temp ^= dfs_Second(y);
               }
           }
 
           // now, checking if xor of subtree
           // computed till now is equal to
           // entire XOR of tree (root)
           // i.e. value at cal_XOR[0] ->
           // which will give entire XOR of the tree
           if (temp == cal_XOR[0]) {
               // then, make that xor 0
               temp = 0;
 
               // increase count by 1, which
               // means we found a subtree
               counter++;
           }
 
           // return xor of subtree
           // till that point
           return temp;
       }
        
       // Function to add edges
       function addEdge(u, v) {
           adj[u].push(v);
           adj[v].push(u);
       }
        
       // Function to take input
       // for (n-1) edges
       function init(edges, numEdges) {
           for (let i = 0; i < numEdges; i++) {
               edges[i][0]--;
               edges[i][1]--;
               addEdge(edges[i][0], edges[i][1]);
           }
       }
 
       // Driver Code
       // taking input
       let n = 5, p = 2;
       nodes[0] = 1, nodes[1] = 6, nodes[2] = 4,
           nodes[3] = 1, nodes[4] = 2;
 
       // making our visited array false
       for (let i = 0; i < n; i++) {
           visited[i] = false;
       }
 
       // taking input for (n-1) edges
       let edges =
           [[1, 2], [2, 3], [1, 4], [4, 5]];
       init(edges, 4);
 
       // First dfs Function call
       dfs_First(0);
 
       // again, marking visited array to false
       for (let i = 0; i < n; i++) {
           visited[i] = false;
       }
 
       // initializing answer variable
       let answer = false;
 
       // if we found XOR of entire tree
       // equal to zero, answer = "YES", as
       // it means there are two connected
       // components equal to each other
       // thus, a single edge can be removed
       if (cal_XOR[0] == 0) {
           answer = true;
       }
       else {
           // second DFS function call
           dfs_Second(0);
 
           // if we found 2 subtree having
           // equal value with XOR of entire tree
           // answer is always "YES", such that
           // p > 2
           if (counter >= 2 && p != 2) {
               answer = true;
           }
       }
        
       // printing the final answer
       if (answer == true) {
           document.write("YES" + '<br>')
       }
       else {
           document.write("NO" + '<br>')
       }
 
       // making counter = 0 for next iteration
       counter = 0;
 
       for (let i = 0; i < n; i++) {
 
           // similarly clearing adjacency list
           // and both arr and cal_XOR array
           adj[i] = [];
           cal_XOR[i] = nodes[i] = 0;
       }
 
       // This code is contributed by Potta Lokesh
   

输出

YES

时间复杂度： O(N+E)，其中 N= 节点数，E = 边数
辅助空间： O(N+E)，其中 N= 节点数，E = 边数